How do you solve the equation $log_2(y+2)-log_2(y-2)=1$ ?

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Answer 1 · 2016-10-23T02:12:27

$y=6$

$log_2(y+2)-log_2(y-2)=1$

We can combine the two terms on the left side:

$log_2((y+2)/(y-2))=1$

We can now deal with the log:

$2^(log_2((y+2)/(y-2)))=2^1$

$(y+2)/(y-2)=2$

$(y+2)=2(y-2)$

$y+2=2y-4$

$6=y$

And let's check it (which can also help get more comfortable with log operations):

$log_2(y+2)-log_2(y-2)=1$

$log_2(6+2)-log_2(6-2)=1$

$log_2(8)-log_2(4)=1$

$3-2=1$

$1=1$

How do you solve the equation log_2(y+2)-log_2(y-2)=1log_2(y+2)-log_2(y-2)=1?